Question
The average weight of five friends A, B, C, D and E is
(x+5) years while the average weight of C and E is (x-5) kg. If the weight of another person F is also added, then average weight of all of them is reduced by 6 kg. Find the value of ‘x’ if average weight of A, B, D and F is 96 kg.Solution
Total weight of friends A, B, C, D and E = 5(x+5) kg So, Total weight of A, B and D = 5(x+5) – 2(x-5) = (3x+35) kg Weight of F = [(x+5-6)×6] - [5(x+5)] => F = (x-31) kg According to the question, => (3x+35) + (x-31) = 96 × 4 => 4x + 4 = 384 => 4x = 380 => x = 95
(22 × 52 ) + 4 × 6 = ? - √324
What should come in place of (?) question mark in the given expression.
 (25% of 320) + (3/8 of 400) − 30 = ?
(5832)1/3  × 10.11 × 11.97 ÷ 16.32 = ? + 45.022
82% of 400 + √(?) = 130% of 600 - 85% of 400
If (x + 1/x) = 5, then value of x3 + 1/x3 is:
Simplify: (1 ÷ 0.08)
What should come in place of (?) question mark in the given expression.
{ (144 ÷ 12) × 5 } − (18 ÷ 3) = ?
Simplify the following expressions and choose the correct option.
(3/4 of 256) + (2/5 of 150) - (72 ÷ 7)
464 + 181 +? = (154 × 25) - (15) 2 Â
15% of 1800 + 22 = ?Â
